A monoatomic gas expands isobarically. The percentage of heat supplied that increases the thermal en — Thermodynamics and Thermochemistry Chemistry Question

💡 Solution & Explanation

Heat supplied $q = n C_p \Delta T$. Increase in thermal energy $\Delta U = n C_v \Delta T$. Work done $= n R \Delta T$. For a monoatomic ideal gas, $C_v = 3/2 R$ and $C_p = 5/2 R$. Fraction for thermal energy = $C_v / C_p = (3/2) / (5/2) = 3/5 = 60\%$. Fraction for work = $R / C_p = 1 / (5/2) = 2/5 = 40\%$. The ratio is 60:40.